Word Problem to find length of Rectangle side using Trig

[edawb]

I solved this exercise problem yesterday. It took me a long while to solve it. Today I tried again and I have no idea how I solved it. I erased my solution from my whiteboard.

Here is the exercise: PQRS is a rectangle. A semicircle drawn with PQ as diameter cuts RS at A and B. The length PQ is 10 cm, and angle BQP is 30°. Calculate the length of PS.

I'm guessing my first problem may be not accurately drawing the diagram. Below is the diagram I'm starting with as it seems to make the most sense. PS is equal to QR so given that I have a triangle with base QR I would try to find the length of QR. Since BQP is 30° BQR should be 60°. And so [imath]QR = \frac{RB}{tan60°}[/imath] or [imath]QR = \frac{10-BS}{tan60°}[/imath]

Thanks in advance.

 

[edawb]

My wife found this solution online. I guess she's a better googler than me. I'm posting this for any other poor soul who comes across this problem. But for the record I solved this but not using this solution. If I can remember how I did it. I'll update this post.

 

[jonah2.0]

Beer induced query follows.

My wife found this solution online. I guess she's a better googler than me. I'm posting this for any other poor soul who comes across this problem. But for the record I solved this but not using this solution. If I can remember how I did it. I'll update this post.

View attachment 34113

Have you remembered it yet?

 

[edawb]

I have not. I'm going to continue until I figure it out. The issue is the book I'm working from does not assume the reader knows about the rule that was provided in the solution I posted.

The book only assumes you know SOH CAH TOA, Pythagora's theorem, the sin, cos, tan of 45° & sin, cos, tan of 30° & 60° as well as basic projections. So only armed with that knowledge one is supposed to be able to solve this problem.

 

[jonah2.0]

Beer induced suggestion follows.

I have not. I'm going to continue until I figure it out. The issue is the book I'm working from does not assume the reader knows about the rule that was provided in the solution I posted.

The book only assumes you know SOH CAH TOA, Pythagora's theorem, the sin, cos, tan of 45° & sin, cos, tan of 30° & 60° as well as basic projections. So only armed with that knowledge one is supposed to be able to solve this problem.

You could also consider the law of sines by drawing a line from point P to B; thus triangle PBQ.

 

[Dr.Peterson]

I have not. I'm going to continue until I figure it out. The issue is the book I'm working from does not assume the reader knows about the rule that was provided in the solution I posted.

The book only assumes you know SOH CAH TOA, Pythagora's theorem, the sin, cos, tan of 45° & sin, cos, tan of 30° & 60° as well as basic projections. So only armed with that knowledge one is supposed to be able to solve this problem.

What you'll probably need to do is to draw in the radius to B and apply Pythagoras to the resulting right triangle. If you let BS = y, you can then have two equations in x and y, and solve for them.

 

[edawb]

Beer induced suggestion follows.

You could also consider the law of sines by drawing a line from point P to B; thus triangle PBQ.

Law of sines hasn't been covered yet. I looked it up and indeed can solve that way so thank you. I learned something new.

 

[edawb]

What you'll probably need to do is to draw in the radius to B and apply Pythagoras to the resulting right triangle. If you let BS = y, you can then have two equations in x and y, and solve for them.

I'm sorry I'm not quite following. But this does sound like how I originally solved this problem but at this point I just don't remember my solution anymore.

When you say draw in the radius do you mean a line from the midpoint of PQ to point B? If so what resulting right triangle are talking about? Should it look like the image in second diagram I posted?

 

[Dr.Peterson]

I'm sorry I'm not quite following. But this does sound like how I originally solved this problem but at this point I just don't remember my solution anymore.

When you say draw in the radius do you mean a line from the midpoint of PQ to point B? If so what resulting right triangle are talking about? Should it look like the image in second diagram I posted?

Yes, I'm talking about this:

​

The triangle I referred to is BCD, where you know the hypotenuse, and can write expressions in x and y for the legs, without having to know about the 60 degree angle.

The other equation comes from the 30 degree triangle QBD (or QBR).

 

[The Highlander]

I solved this exercise problem yesterday. It took me a long while to solve it. Today I tried again and I have no idea how I solved it. I erased my solution from my whiteboard.

Here is the exercise: PQRS is a rectangle. A semicircle drawn with PQ as diameter cuts RS at A and B. The length PQ is 10 cm, and angle BQP is 30°. Calculate the length of PS.

I'm guessing my first problem may be not accurately drawing the diagram. Below is the diagram I'm starting with as it seems to make the most sense. PS is equal to QR so given that I have a triangle with base QR I would try to find the length of QR. Since BQP is 30° BQR should be 60°. And so [imath]QR = \frac{RB}{tan60°}[/imath] or [imath]QR = \frac{10-BS}{tan60°}[/imath]

Thanks in advance.

View attachment 34112

I have not. I'm going to continue until I figure it out. The issue is the book I'm working from does not assume the reader knows about the rule that was provided in the solution I posted.

The book only assumes you know SOH CAH TOA, Pythagora's theorem, the sin, cos, tan of 45° & sin, cos, tan of 30° & 60° as well as basic projections. So only armed with that knowledge one is supposed to be able to solve this problem.

Hi,

I’m not entirely sure where your difficulties have come from (except that you do appear to be flying off into Tangents…unnecessarily!) and your green sketch doesn’t appear to reflect the situation accurately.

Did the original question showing that “5 cm dimension” in a not particularly "useful" position somehow throw you?

If you start with the original figure and then follow the “instructions/hints” given in the original text, then you should end up with something more like this:-
(Rather than the second figure shown in the original text)

​

It should now be obvious (I trust) that:-

\(\displaystyle Sine (60°) = \frac{DB}{5} \Rightarrow \frac{\sqrt{3}}{2}=\frac{DB}{5} \Rightarrow \frac{\sqrt{3}×5}{2}=DB \Rightarrow DB\approx4.33cm\) (to 2 d.p.)

And, since PS = DB, then PS ≈ 4.33cm too (or PS = \(\displaystyle 2.5\sqrt{3}\)cm).

Alternatively, using “knowledge” of "the ratios of a 30-60-90 triangle", you have:-

​

And since the hypotenuse (BC) is the “5cm” side you would simply divide that by 2 (incidentally giving the length of the shorter side (CD) = 2.5cm) and then multiply that answer by √3 to get the length of DB &, hence, that of PS.

This is really quite a simple problem (if approached correctly). I’m surprised nobody else has "explained" it already. ?

 

[Dr.Peterson]

Did the original question showing that “5 cm dimension” in a not particularly "useful" position somehow throw you?

As I understand it, that picture is not from the original question, but from a website they found giving a solution they couldn't use, not having learned the theorem about inscribed angles.

If you start with the original figure and then follow the “instructions/hints” given in the original text, then you should end up with something more like this:-
(Rather than the second figure shown in the original text)

Again, that's not from the original text, which appears not to have included a figure.

It should now be obvious (I trust) that:-

\(\displaystyle Sine (60°) = \frac{DB}{5} \Rightarrow \frac{\sqrt{3}}{2}=\frac{DB}{5} \Rightarrow \frac{\sqrt{3}×5}{2}=DB \Rightarrow DB\approx4.33cm\) (to 2 d.p.)

But he said,

The issue is the book I'm working from does not assume the reader knows about the rule that was provided in the solution I posted.

The book only assumes you know SOH CAH TOA, Pythagora's theorem, the sin, cos, tan of 45° & sin, cos, tan of 30° & 60° as well as basic projections. So only armed with that knowledge one is supposed to be able to solve this problem.

So what I have been helping with is his own approach, which does not use the 60° angle that your approach depends on. Yes, that fact makes the problem easy; but it's good to help a student use what he knows. I like to work with a student, rather than tell him what he should do instead, when he is not doing anything actually wrong.